[英]Create table fails if foreign keys are used in SQLite
我真的不知道为什么这个 php 代码没有创建三个表。 第一个和第二个都很好。 但是第三个失败了。 第三个与第二个非常相似,但在外键定义之后有另一个字段。 我在这里做错了什么?
$db = new \PDO("sqlite:d:/temp/test_db.sqlite");
$db->exec("PRAGMA foreign_keys = 'ON'");
$statement = $db->query("CREATE TABLE IF NOT EXISTS customers(
id TEXT PRIMARY KEY NOT NULL ,
name TEXT
)");
$statement->execute();
$statement = $db->query("CREATE TABLE IF NOT EXISTS appointments (
id TEXT PRIMARY KEY NOT NULL ,
customer TEXT ,
FOREIGN KEY (customer) REFERENCES customers(id)
)");
$statement->execute();
$statement = $db->query("CREATE TABLE IF NOT EXISTS appointment (
id TEXT PRIMARY KEY NOT NULL ,
customer TEXT ,
FOREIGN KEY (customer) REFERENCES customers(id),
nextfield TEXT
)");
$statement->execute();
根据SQLite语法规范, nextfield TEXT
应与其他列定义放在一起。 FOREIGN KEY
与table-constraint
部分有关,应在列定义之后定义:
$statement = $db->query("CREATE TABLE IF NOT EXISTS appointment (
id TEXT PRIMARY KEY NOT NULL ,
customer TEXT ,
nextfield TEXT,
FOREIGN KEY (customer) REFERENCES customers(id)
)");
FOREIGN KEY必须在列之后编码,或者作为列定义的一部分而不用,FOREIGN KEY(列)作为隐含的列,因此替换
"CREATE TABLE IF NOT EXISTS appointment (
id TEXT PRIMARY KEY NOT NULL ,
customer TEXT ,
FOREIGN KEY (customer) REFERENCES customers(id),
nextfield TEXT
)"
with(表约束): -
"CREATE TABLE IF NOT EXISTS appointment (
id TEXT PRIMARY KEY NOT NULL ,
customer TEXT ,
nextfield TEXT ,
FOREIGN KEY (customer) REFERENCES customers(id))"
或者(列约束): -
"CREATE TABLE IF NOT EXISTS appointment (
id TEXT PRIMARY KEY NOT NULL ,
customer TEXT REFERENCES customers(id) ,
nextfield TEXT)"
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.